Mixing
Multivariable integration $int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$begingroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
602313
$endgroup$
add a comment |
$begingroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
602313
$endgroup$
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
add a comment |
$begingroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
602313
$endgroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
602313
asked Jan 13 at 11:56
LeifLeif
602313
edited Jan 13 at 12:35
Leif
asked Jan 13 at 11:56
LeifLeif
602313
asked Jan 13 at 11:56
LeifLeif
602313
asked Jan 13 at 11:56
LeifLeif
602313
602313
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
add a comment |
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
add a comment |
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$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54